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📐 Mathematics  ·  Circles & Mensuration  ·  JEE

In a circle of radius R, two chords AB and CD intersect at P. Then PA × PB =

Answer: PC × PD.

  • A PC × PD
  • B (PC + PD)/2
  • C PC² + PD²
  • D PC - PD

Correct answer: A. PC × PD

Explanation: Intersecting chords theorem: triangles PAC and PDB are similar (angles in same segment). So PA/PC=PD/PB, giving PA×PB=PC×PD. This product equals the absolute value of the power of point P w.r.t. the circle. Ans: PC×PD.

Cylinder, Cone, SphereV=πr²hCSA=2πrhV=⅓πr²hCSA=πrlV=4/3πr³SA=4πr²

Three standard 3D solids and their key formulas: a cylinder's volume scales with the full height, a cone's volume is exactly one-third of the cylinder with the same base and height, and a sphere's volume and surface area both follow distinct r-based formulas worth memorising separately.

Concept context

Perimeters, areas, and volumes of 2D and 3D shapes

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